# Monty Hall, Monty Fall, Monty Crawl

Good lord - I can’t believe you losers are still debating this. There is SO much literature and precedent on this question you are not going to solve it here.

The people who say 50% are wrong. The people who say 66% are right.

Try to understand.

Even if you don’t understand, the answer is still the same.

Ask yourself, the car doesn’t magically move once that non-winning door is opened, now does it? When all the doors were closed, and you picked one, you had a 1-in-3 chance of getting the car. That means there’s a 2-in-3 chance the car is SOMEWHERE ELSE. So when the host opens one door, revealing a goat, that means that 2-in-3 probability has focused on the Other Door that you didn’t choose.

If you play with a million doors, the odds the car is behind the last unopened door are not 1/2, they are 999,999-in-1,000,000! The car itself does not shift around, only the probabilities do.

Jeff, in the other article, the problem you stated was nothing like the Monty Hall problem. In this one, it’s 66% because Monty Hall cannot open a door that has the car. In the other, it’s 50% because we only need to know what the other child is.

The way I think of it is the only way I can lose by switching is if I picked correctly the first time which was 1 in 3 odds. If I picked either incorrect box switching means I win since the host must reveal an empty box.

The combinations of boy and girl are a red herring. They’re statistically meaningless unless you can get everyone on the planet to agree to having two children.

The sex of a child is always independent of the sex of other children.

BA, Yes of course the sex of one child is independent of the other children. We know that. Everybody knows that.

Doesn’t answer the question though. You’ve fallen for the red herring and completely skipped the question. Sucker.

WWJD = What Would Jesus Do.

Actually, it’s a good messiah interview question. That and get the solutions to np-hard problems. Oh and while you’re at it, ask if the moon landing was a hoax or not.

Man I love omniscience.

This blows my mind.

So what’s the deal with the real Monty game anyways?

1. Does Monty always reduce the number of doors from 3 to 2?
2. Does Monty never open the door to the car in 1. ?
3. Does Monty have a tell, and did the outcome of the real Monty Hall games agree with statistical simulations?

"Good lord - I can’t believe you losers are still debating this. "

Amen. What is it about Jeff’s “it’s been covered to death” don’t you people get. If you want to debate it, go somewhere else. Better yet, just google it and read up on it and save yourself the headache. There is nothing to debate about this problem.

Ok then, what’s the question? I thought it was “What are the chances that the other child is CHOSEN_SEX?”.

By concluding that all combinations of of boy and girl are equally likely you are making a statistical fallacy in assuming that a randomly selected subset has the same properties as the superset.

To illustrate, pretend we have a planet where people can choose to have either one child or two children (for whatever reason, the point is to reduce the sample space). And lets say that 2/3 the population chooses to have one child. Now, by some random chance, all of the people who chose to have one child all had boys. Statistically, that just ruins everything. Because in this scenario, if you have two children, it is more likely that you have two girls.

Why? Because half of all births are already boys. Assuming an even distribution of boys and girl births, we can expect two child couples to have only girls.

To use the general case that it is 50% likely for a given sex during birth as true for all subsets is wrong. Especially since we aren’t scooping up a random subset, but a subset that also exhibits an additional specific quality.

So, in conclusion, we honestly cannot know with any degree of certainty the sex of a given child based on the sex of another child, no matter how you finagle the language. There are simply too many other variables.

It’s also the problem with the “proofs” provided. They all just assume two child families and run data using two child sets without thinking about the likelihood of having a two child set in the first place.

Data from people I know:
My parents: GBGB
My older sister: G
Her husband’s parents: BGB
My wife’s parents: BGG
My younger sister: GGB
My aunt and uncle: GGB
My boss: B
My coworker: B
My step-father: GB
His daughter: GG
My sister’s neighbor: BB

13 girls, 12 boys, 3 two child families out of 11. Not a terribly large sample size, true, but it is representative of the inherent instability of what the question is asking. The people who are claiming 66% are claiming to know more information than than actually have available. The event is so definitely random that the best thing to do is to flip a coin and call it 50%. Anything else is being dishonest on some level.

"The event is so definitely random that the best thing to do is to flip a coin and call it 50%. "

And that’s exactly what many of us proposed that you do to prove the final conclusion to yourself. And yet, you obviously haven’t done that!

I really don’t know why people keep arguing around in circles on these things instead of just doing a quick simulation with cards or a coin and figuring out the REAL answer. It’s so much easier and you don’t have to even think about it.

Sounds like a fairly standard application of Bayes’ Theorem.

I could be wrong, but I get the same answer as everyone that resorted to counting or simulations.

Just like a bunch of programmers to go back to basic counting and complicated simulations rather than use an existing, well tested algorithm.

The linked article is ridiculous. A made up “Proportionality Principle” claiming to be a restatement of Bayes’ Theorem, when the Wiki entry on Bayes’ theorem actually uses Monty Hall as an example and gets the right answer.

Probability is a measure of information. People getting hung up on the “probability that an event occurred” without realizing that it doesn’t matter. We know the event occurred.

People can make their own red/black cards as stated in the pdff. They can sit there and count how many times they flip the red card over and see red on the back. It will happen 1/2 the time (unlike the Monty Hall problem). You can play with a Monty Hall simulator online. Try it 100 times without switching, you’ll win about 33 times. Try it 100 times switching every time, you’ll win about 66 times. Sheesh.

@BA:

The problem explicitly states that they have two children. So trying to take into account the number of children other families have is a wrong assumption. Plus, even if it was correct, two children out of a 7 billion people isn’t going to make, statistically, much of a difference. Anyway.

I will note, however, that there is an assumption that the chance of a child being a boy or a girl is 50% each way.

The four possible combinations of children would be BB, BG, GB, and GG. We know there is at least one girl, so BB is impossible. Therefore, the only choices that are left are BG, GB, and GG. Which would say the choices for the other child are B, B, and G. There is a 1/3 chance of the other child being a girl.

It’s not a classic case of “P(A) Given B,” because A is a subproblem of B, not the other way around. The problem you’re used to is “I’ve already had a child who is a girl, so what’s the odds of my next child, which I haven’t given birth to, being a girl?” This problem is “I’ve already had two children. One of those children is a girl; what’s the probability the other one is as well?”

And I’d think, if you looked at a large group of people with two children, and you selected families with at least one girl, you’d see the same thing: 1/3 have two female children, 2/3 have a girl and a boy.

Then let’s define some parameters and get some basic data.

What are the likelihoods for the following:

1 child families
2 child families
3 child families
4 child families
5 child families
6 +child families

Basically the algorithm will first determine the family size based upon the previous likelihoods. Then generate enough children to populate the family. Then we discard the family if it does not have two children.
Then we can get the statistics for all the different combinations and run it through the proper formula to find the ending result.
Repeat this for some sufficiently large number and aggregate the results to find the most likely answer within a standard deviation.
(Of course, the algorithm should randomly start new families and generate children and randomly deposit them into all active families instead of filling them in as they come, but it’s a cut I can live with.)

Another way to go about it is to generate a sufficiently large number of children and divide them among the different sized families according to the size probabilities. And then just use the two child families.

That’s my biggest problem with all of the mathematical and programmatic proofs provided. They all seem to be ignoring the fact that not all families have two children and the distribution among the different combinations may not be even.

In other words, why is it safe to ignore all the other children in the world when considering this problem?

@Birch

But to treat the problem as simple combinations is to exclude a bunch of relevant data as I tried to show in my hypothetical world. Yes, in isolation the probabilities of each set is 25% and we can exclude one set because of what we know currently and each remaining set has equal probability which would then bring them all to 33% of which two of our choices are positives which would then give us a 67% chance positive.

Let me state that I understand that. Perfectly. In a world of two child families, this would be true 100% of the time.

But the problem does not stand in isolation from the world. Not statistically. Because the statistics of them being a two-child family is a consideration itself. Plus the ratio of girls to boys in non-2 child families would affect the ratio of girls to boys in 2 child families given that the ratio of all girls to all boys is 1:1.

I’m questioning the assumption of treating 2 child families as if they represented the whole.

Will people who use the Monty Hall problem in interviews remember to explain Mr Hall’s motivation (is he trying to make you win/lose).

Otherwise the question is unanswerable - it might have been useful 200years ago when the show was on TV, but now when people read it 10th hand out of a interview techniques book it doesn’t make sense.

Quite interesting question, and debates thereafter.
That reminds me of Nicholas Nassim Taleb, who wrote two books “Fooled by Randomness” and “The Black Swan”, and one of the themes in both the books is that human mind isnt really attuned to understand randomness(though this summary is perhaps an over-simplification).

@just jeff
"Ask yourself, the car doesn’t magically move once that non-winning door is opened, now does it? When all the doors were closed, and you picked one, you had a 1-in-3 chance of getting the car. That means there’s a 2-in-3 chance the car is SOMEWHERE ELSE. So when the host opens one door, revealing a goat, that means that 2-in-3 probability has focused on the Other Door that you didn’t choose."
In other words, there is higher chance of choosing the wrong door than right one the first time it is choosen.
Actually, IMO there’s actually a way to solve and resolve this Monty Hall Problem once and for all: Good old Empiricism.
Find lots of people or record high number of observations(high number need due to Law of Large Number in statistics) where the game of guessing which one of the three doors the object is behind. If it happens statistically that more people have gotten it wrong than right the first time they choose the door, then this proves that at least changing the door is a good choice.

The first time I came across this problem was in “The Curious Incident of the Dog in the Night-time,” and I thought it was explained very well there. I like the explanations above also to think of a million doors and narrowing down from there.